VIEWS: 5 PAGES: 19 POSTED ON: 6/28/2012 Public Domain
Two Full Solutions for a Simple RC Network Steve Keith http://www.baselines.com This paper will show how to find the voltage across a network containing a current source, a resistor and a capacitor in parallel. The current source will be a pulse of width T. The first method will show a time domain analysis, and the second method will show a frequency domain analysis that demonstrates how to use residues and Fourier transforms. 1 – The circuit and the current source time graph 2 – Time Domain Analysis The current analysis will be treated in two steps, thanks to superposition; a positive unit step at 0 and a negative unit step at T. Assume that the circuit has no initial conditions before time=0. Some equations to remember (v = Voltage, i=Current, R = Resistance and C = Capacitance: Ohm’s Law for a resistor and an extension of Ohm’s Law for a Capacitor KCL states that all of the currents entering and leaving a node must equal zero, so the current coming into the top node has to equal the sum of the two currents leaving that node. A few things to remember: - The voltage across a capacitor cannot change instantaneously. - A capacitor is like an open circuit to a signal that has been the same for a long time (DC) - The current through a capacitor can change instantaneously - t(0+) means the time as close to zero in the positive direction as possible. - t(0-) means the time as close to zero in the negative direction as possible. At t(0-) the current has not yet begun to flow, and since the voltage across the capacitor cannot instantaneously change, then at t(0+) the voltage is still 0. The first thing we will do is examine the natural, or source free response of this circuit. To do that, we look at the circuit at t(0-), before the current is connected to the circuit. We set i(t) equal to zero in the equation above, and find that: After a little algebraic rearranging it turns out that: Integrating both sides (t goes from 0 to t, v goes from v(0+) to v(t)): Taking the antilog of both sides, rearranging and renaming v(0+) to : This is the form of the natural response, an exponentially decaying function that starts at the initial value of the voltage and decays to zero. The rate of decay is dependent on the values of the resistor and the capacitor. Here is a sample plot. 5 4 3 2 1 1 2 3 4 5 This shows that if a circuit of this type has a voltage across the capacitor at time t(0+), and there are no active sources, this voltage will decay over time. The capacitor gives up its charge by ‘bleeding’ through the resistor. The charge is lost due to the generation and release of thermal energy. Now that we know the natural (source free) response of the circuit, we are ready to take a look at what happens when a source is applied. Using intuitive reasoning, we would assume that the voltage across the capacitor would initially be zero and that it would ‘ramp-up’ exponentially to a maximum value that it will maintain until the source is removed from the circuit. When that happens, the capacitor voltage would exponentially decay to zero, bleeding its energy through the resistor. In fact, that is what happens. To determine the voltage response, we will use superposition to split the problem into two parts. We will calculate the response for the rise of the current from 0 to 1 at t=0 and maintaining a value of 1 until t=infinity. Then we will calculate the response for a drop in current from 0 to -1 at t=T, and maintaining a value of -1 until t=infinity. Adding these two together gives us our pulse as defined in the first section. See the figure below. If we look at our circuit, and assume that the source is always on from t=0 to t=infinity, we can calculate what the final value of voltage will be at t=infinity. Knowing that a capacitor acts like an open circuit to a dc signal, we know that the voltage drop across the resistor due to current I is R*I. In our case, we know that the current attains a maximum value of 1, so the maximum voltage is equal to R. We also know that the circuit must act in accordance with the natural response equation we determined above. Here is where initial and final conditions can help us out. We know that when t=0+, v(t) must be 0, and we also know that at t=infinity, v(t) must be equal to R. In order to comply with these results and still use our natural response equation, some inspection must be done. Plugging these time and voltage values into the natural response equation and after some thought, we can determine that the full response must be the following: or Try plugging 0 and infinity into this equation and you will see that at 0, v(t) = 0 and at infinity it equals R. See the sample chart below for this function. It shows the voltage exponentially increasing from 0 to R volts. For this sample chart, R=5. 4 3 2 1 1 2 3 4 5 So this is the first part of our problem solved. Now we must find out what happens when the input current is a transition in the negative direction (i.e. from 0 to -1) at t=0+. After we do this, we can adjust the timing (slide negative pulse to the right) so that this negative pulse occurs at t=T and use the value of our voltage at T (from the increasing exponential term) due to the positive pulse we just calculated. Superposition tells us we can then add the two signals together to get the desired voltage response to the complete current pulse. Let’s start from scratch and have a negative step of 1 be applied at t=0. Forget about the positive step we just calculated for now. When this equation is manipulated, as above, we attain the same result for natural response, which we would expect since the values of R and C haven’t changed. However, in this case, is a negative number, because of the polarity we chose for v(t). Again, using our intuition, we know that the voltage at t(0+) will not change immediately, so it starts at 0. At t = infinity, the voltage would approach –R, and the approach is exponential as dictated by the values of R and C. In order for these conditions to be met, we again need to inspect our natural response and tailor it to fit the situation. We find that: or We are not done yet. We need to adjust this so that the negative pulse happens at t=T and not at t=0. To do this, we need to adjust the exponential value of t as shown below. Testing this out, we see that at t=T, the voltage is 0 and at t=infinity, the voltage is –R as we wish. At this point, we have solved the two pulse responses and we need to add them together for the period where they are both active (t>=T) as superposition dictates, to get the full response. t>=T We can simplify this equation. Below are the two results, one is the voltage response during the current pulse, and one is the voltage response after the pulse. Full Time Domain Solution: This is the solution for the voltage response. Here’s a real world example. Resistor = 5KΩ Capacitor=0.5µf Current= 1 Amp Pulse of Period 7ms 3 – The frequency Domain Analysis using Residues and Fourier Transforms We are going to use the same circuit we used before, and the same current pulse as a source. Here is our strategy of attack. 1 – Write the circuit equation in the time domain and convert it to frequency domain. 2 – Fourier transform of the time domain of the supplied current into the frequency domain. 3 – Calculation of the frequency domain voltage from the frequency domain current. 4 – Conversion of frequency domain voltage to time domain voltage using residues. 5 – Compare with time domain answers. This becomes much more mathematical than physical. To gain a feeling for what is going on here, realize that even though we are dealing with physical circuits, we are translating them into ideal ‘mind circuits’ with currents and voltages related to other circuit elements being just mathematical functions. The variables dealt with when using these mathematical functions, to more widely encompass real world situations can digress fully in complex math, which deals with imaginary numbers. To that point, the voltage and current functions are examined for all possible values that could be plugged into the variables. This means we will be using the complex plane where the x-axis contains real values, the y-axis contains imaginary values and all other areas of the plane are a combination of both real and imaginary (eg 1+i4). To capture all values, the functions are evaluated along the x axis as a function of x, and in a semicircular area in either the upper half plane or the lower half plane as a function of z, the complex variable (eg z=x+iy). This semicircular area is allowed to go to infinity. Any singularities (poles or zeroes) that cannot be removed must be accounted for. If there happen to be singularities along the x axis that cannot be removed, then the path of integration must go around them. With the help of certain rules, we can shrink the semi-circle around these singularities on the x axis down by letting the radius go to zero. This will be demonstrated in this example. ==================================================================================== First some needed formulas and concepts; those are not derived, but merely stated, here. FOURIER TRANSFORM PAIR – These two equations transform a time domain signal ( ) to a frequency domain signal ( . A major reason for using the transform is that it can make problems that are intractable or difficult in the time domain much easier to compute in the frequency domain. Concept 1: In the time domain, if you have a derivative term, it transforms into a multiplication by the complex frequency ( ) in the frequency domain. So here is what a capacitor current transforms into: Time Domain =======Frequency Domain ======= Here the i in is the imaginary number . Concept 2: Cauchy-Gorsat Theorem – If the function you are working with is analytic (no singularities) in any closed contour in the z-plane (the complex plane), then the value of the integral of that function around the contour is zero. Concept 3: L’Hopital’s Rule – This will be useful to us in determining whether a singularity is removeable or not. If you have a function that has a numerator [g(z)] and a denominator [h(z)], and at some point, , both the numerator and denominator are zero, and if both numerator and denominator are differentiable at this point, and , then (The apostrophe means the derivative of). The benefit here is that if the function is undefined (eg zero/zero) at a location, there may be a non- infinite value, which can be found by taking the derivative of the numerator and denominator and calculating the value of that. If this produces a non-infinite number, then the singularity at that point is removable. Concept 4: Jordan’s Lemma – for any rational function P(z) / Q(z), if the degree of the polynomial Q(z) exceeds that of P(z) by at least one, and if the exponent of the e term below is positive, then the integral around a semicircular arc contour in the upper half complex plane is zero: r is the radius of the semicircular contour in the upper half plane. V>0. There is a similar rule for when v<0, in which case, the semicircular area in the lower half plane is used, and the direction of the contour integration is reversed. Concept 5: Residues – If a function f(z) has only one singularity at , and you enclose this singularity within a closed contour in the complex plane, then the residue can be calculated with the following formula. There are a number of ways to find residues. Residues are useful in determining what the integral of the function f(z) is. In some cases, using residues is the only way to integrate difficult functions. More details on deriving formulas useful in residue calculus are beyond the scope of this paper, but one more formula is provided below that will be helpful in our example: at the point . If there are more than one singularity, you calculate the residue at each singularity and add them together. Multiplying by above will eliminate the singularity in the denominator, as will be demonstrated later. Concept 6: Dealing with singularities on the real axis (Integrals involving indented contours). If f(z) has a simple pole at , and an arc is drawn with this singularity as the focal point, and the arc subtends an angle of α radians, then The integration is done in the counter-clockwise direction. (If the integration is done in the clockwise direction, put a negative sign in front of the equation above.) This formula is useful for dealing with singularities on the x axis as will be demonstrated later. =================================================================================== The information provided above should be enough to allow us to continue our discussion of the RC circuit presented in the time domain section, repeated here for convenience. The gory details We start with the same time domain equation that we wrote for the circuit earlier and convert it into the frequency domain equation using concept 1. Time Domain ==========Frequency Domain ========== After rearranging terms we see that: The Fourier transform for the current as a function of frequency (from the equations above) is: Since the current is a pulse of 1 unit between 0 and T, we can confine the limits and rewrite this equation as: The integral evaluates to this: Plugging this into the equation above for Using the Fourier Transform to move from Frequency to Time Domain: Here is where the usage of residues is useful in evaluating the integral above. The first step is to determine the singularities in this horrendous looking equation. Let’s look at the first term after some algebraic manipulation: So you see that when this term blows up! This is one of our singularities. Next we see that there is an omega ( in the denominator. However, looking at the equation as a whole, when you substitute in the numerator, then the numerator and the denominator both go to zero. This is where we need concept 3, L’Hopital’s Rule. In our case, g(z), the numerator is And the h(z), the denominator is We need to take the derivative of these two equations (with respect to omega), plug in the value , and calculate the value of the whole function at that point (the singularity). If we can do that and get a non-infinite, non-zero value, we can remove this singularity from the calculation. The derivatives are easy to calculate and you can also use Wolfram Alpha (interface available at http://www.baselines.com/) to do the derivatives for you. The derivative of the numerator is The derivative of the denominator is When 0 is substituted for omega, this reduces to This is the value of our function when =0. Since it is not an infinite value, we can remove this singularity from our calculations. The next step is to calculate the residue at the remaining singularity Lets draw out the situation in the complex plane here, showing the contour path and the singularity: The singularity i/RC is right on the Y axis, as it is a pure imaginary number. Here is our equation again, copied from above. We will refer to the integrand as . Concept 5 reformulated for our situation: ] And: So, if we can calculate the residue, we can determine what v(t) is (by multiplying the residue by . You may have noticed that v(t) has integration limits of -∞ to +∞, and that is a contour integral. We capture all values of by letting the radius of the semicircle go to infinity in the contour integration. This allows us to go from -∞ to +∞ in the integration for v(t). We will try to see if this equation is in a format where we can take advantage of Concept 4, Jordan’s lemma. We can see that the numerator has a polynomial of order 1, and that the denominator has a polynomial of order 2 ( . In addition, if t > T, we can be assured that the exponential term is positive. Thus, the two conditions in Jordan’s Lemma are met. Given this, the integration around the upper semi- circular arc is zero as r (the radius) goes to infinity, and we only need to worry about the integration around the rest of the closed contour, which is the real axis from –r to +r as r tends toward infinity. We use the upper half plane for the integration to find out what the voltage is during the time when t>T, because this is when the exponential term is positive (Jordan’s Lemma). We will still need to calculate what the voltage is when t<T. We will do that later. For now, when t>T: Using concept 5 : We multiply our by and plug the singularity in for , we get: (Remember Also remember that Again using concept 5, So if we multiply the residue by , we will determine the value of the voltage integral we are looking for. We’ve successfully calculated what v(t) is for t>=t. If you compare this with what we obtained during the time domain portion of this paper, you see they agree. All that is left is to calulate what the voltage will be for the period of time during the current pulse, t>0 and t<=T. This is a bit more work! When t<T, this equation has a negative component in the exponential term, so Jordan’s lemma does not hold for the upper half plane. We will need to use the lower half plane when the exponent goes negative. So lets restate the equation and break it into two simpler parts. Again we call the integrand : One advantage in doing this is that we can use the upper half plane for since the exponent here is positive…this however leads to another complexity which we will deal with next. Now that we have broken the equation apart, we can no longer treat the singularity at as removable. If you plug into both and equations, you see that the numerator becomes 1 and the denominator becomes 0, the sure sign of a pole that cannot be overlooked. We will have to evaluate residues at this pole as well as the one at i/RC in the upper half plane, and then add the two together. Let’s get that part done now before dealing with the lower half plane. In this graph, the light blue area is the closed path we will be integrating. You see the singularity at i/RC as before, and also the new non-removable singularity at 0. The process will be finding the residue at i/RC as before, and then using concept 6 to determine the residue at the origin. Again, to capture the complete solution, we will let the radius of the large semicircle go to infinity and we will allow the radius of the smaller semi-circle to shrink to 0. Concept 5: due to the singularity at i/RC. Concept 6: (Here, r is the radius of the small semicircle. We have a small semicircle around the singularity at zero, so . We combine both factors and we have our answer for The only remaining task is to calculate and use and to calculate total v(t) for t<T. The difficulty here is that when t is less that T, the exponent term is negative, so we must evaluate this function in the lower half complex plane. The one benefit is that the singularity i/RC is no longer in play. It is in the upper half plane. All we need to worry about is the non-removable singularity at The semicircle is a mirror of the one used in the upper half plane. Since this is a mirror image, the direction of integration in the lower half plane is opposite that of the upper half plane, so the result will have a negative sign in front of it (clockwise integration as opposed to counter-clockwise). Since there is no singularity within the large semicircle, all we need to do is evaluate the residue at 0 using concept 6. As shown above, the total solution for the voltage during the current pulse is: Full Residue Solution: Full Time Domain Solution: Thankfully, the solutions agree and we are finished. Time for a beer!